Complex analysis, construct analytic f given Re(f)

[usn7564]

Homework Statement

u(x,y) = sin(x^2-y^2)cosh(2xy)

Find a function f(x+iy) = u(x,y) + iv(x,y), where v(x,y) is a real function, such that f is analytical in all of the complex plane. Find all such f. The attempt at a solution
I expanded using Euler's for sin and cosh which gave me

u(x,y) = \frac{1}{4i} \Bigg( e^{i(x^2-y^2) +2xy} + e^{i(x^2-y^2)-2xy} - e^{-i(x^2-y^2)+2xy} - e^{-i(x^2-y^2)-2xy}\Bigg)
Which I can simplify to
\frac{1}{4i}\Bigg(e^{i\bar{z}^2} - e^{-i\bar{z}^2} + e^{iz^2}-e^{-iz^2}\Bigg)
I don't think the simplification will be necessary to solve it but it shows that u depends on the conjugate of z. This makes me assume v has to have the same terms depending on the conjugate terms, but negative to cancel them out (else f cannot be analytic).
And that's as far as I've come. Simply looking at u(x,y) makes me pretty sure I shouldn't solve it 'conventionally' using the Cauchy-Riemann equations, got to be some trick to it. Though I don't know how to proceed.

The answer should be
v(x,y) = sinh(2xy)cos(x^2-y^2) + C \implies f = sin(z^2) + iC

 

[HallsofIvy]

You don't really need to expand in terms of exponentials. Just use the Riemann-Cauchy equations:
\dfrac{\partial u}{\partial x}= \dfrac{\partial v}{\partial y}
\dfrac{\partial u}{\partial y}= -\dfrac{\partial v}{\partial x}

Here, \dfrac{\partial u}{\partial x}= 2xcos(x^2- y^2)cosh(2xy)+ 2y sin(x^2- y^2)sinh(2xy)
and \dfrac{\partial u}{\partial y}= 2xcos(x^2- y^2)cosh(2xy)+ 2y sin(x^2- y^2)sinh(2xy)

So you must have
\dfrac{\partial v}{\partial x}= 2xcos(x^2- y^2)cosh(2xy)+ 2y sin(x^2- y^2)sinh(2xy)
\dfrac{\partial v}{\partial y= -2xcos(x^2- y^2)cosh(2xy)- 2y sin(x^2- y^2)sinh(2xy)

 

[jackmell]

You don't really need to expand in terms of exponentials. Just use the Riemann-Cauchy equations:

Oh no Hall. You have it wrong. He doesn't feel like muscling-through the Cauchy-Riemann equations. I mean aren't they all suppose to be easy-peasy?

v(x,y)=\int \frac{\partial u}{\partial x} \partial y+\int\left\{-\frac{\partial u}{\partial x}-\frac{\partial}{\partial x}\int \frac{\partial u}{\partial x} \partial y\right\}

Think that's right. May want to check if you like stuff like that (not Hall, the OP I mean).

And I thought it was Dairy Queen.

 

[usn7564]

Oh no Hall. You have it wrong. He doesn't feel like muscling-through the Cauchy-Riemann equations. I mean aren't they all suppose to be easy-peasy?

It's a from a mini-exam that has several questions in 50 min, so yes, they are supposed to be relatively easy. There was another similar question that you could completely trivialize by looking at it for more than a second while solving it conventionally simply wasn't an option.

Maybe the integral doesn't take too long, will take another gander at it.

 

[jackmell]

It's a from a mini-exam that has several questions in 50 min, so yes, they are supposed to be relatively easy. There was another similar question that you could completely trivialize by looking at it for more than a second while solving it conventionally simply wasn't an option.

Maybe the integral doesn't take too long, will take another gander at it.

No, the integral is tedious and messy to use in practice. My point was to emphasize that real-world problems are way more difficult to solve than textbook problems so best to get use to, and not be intimidated by tough problem-solving methods.

The integral is an instrument to test your knowledge about using the Cauchy-Riemann equations to solve your problem. Deriving the integral is I believe a clear indication of mastering that task. So the purpose of the integral is not to use it but rather just to derive it.

Oh yeah, almost forgot: use the C-R equations to solve your problem even if it takes you hours and forget about the mini-exam (my personal recommendation).